Simple Lie groups with exceptionally small automorphism groups Famously
$$
Aut(Spin(8))\cong PSO_8(\mathbb{R})\rtimes S_3 
$$
which is not equal to
$$
Aut(SO_8(\mathbb{R})) \cong PSO_8(\mathbb{R}) \rtimes 2 
$$
see https://math.stackexchange.com/a/3105463/758507
I'm interested in other examples of simple Lie groups whose automorphism group is strictly smaller than the automorphism group of the universal cover.
This cannot happen for types $ B_n, C_n,F_4,G_2,E_7,E_8 $ since every automorphism is inner (the Dynkin diagram has no symmetry so no outerautomorphisms) and thus we groups of those types we always have
$$
Aut(G)=G/Z(G)=Aut(\tilde{G})
$$
If $ G $ is centerless then we have
$$
G \cong \tilde{G}/Z(\tilde{G})
$$
and so, since $ Z(\tilde{G}) $ is a characteristic subgroup of $ \tilde{G} $, we have
$$
Aut(G) \cong Aut(\tilde{G}/Z(\tilde{G}))\cong Aut(\tilde{G})
$$
see https://math.stackexchange.com/a/4038199/758507
So that implies for type $ D_n $
$$
Aut(PSO_{2n}(\mathbb{R}))\cong Aut(Spin_{2n}(\mathbb{R})
$$
for all $ n $. The only other form of $ D_n$ is
$$
SO_{2n}(\mathbb{R})
$$
For $ n\neq 4 $ then the outer automorphism group is cyclic 2 (comes from swapping short legs of Dynkin diagram) and this outer automorphism can be realized by conjugation by $diag(-1,1,\dots, 1)\in O_{2n}(\mathbb{R})  $. Thus
$$
Aut(SO_{2n}(\mathbb{R}))\cong Aut(Spin_{2n}(\mathbb{R})
$$
for $ n \neq 4 $. $ SO_8 $ has exceptionally small automorphism group, as noted above.
What about the $ A_{n-1} $ series? Again we know
$$
Aut(PU_{n})\cong Aut(SU_n)
$$
what about intermediate groups
$$
SU_n/\zeta_d
$$
for some $ d $ dividing $ n $ (here $ \zeta_d $ is a primitve $ d $th root of unity). Does $ SU_n/\zeta_d $ ever have strictly smaller automorphism group? Seems probably not since the only outer automorphism of $ SU_n $ is complex conjugation (flipping Dynkin diagram) and complex conjugation preserves the group generated by $ \zeta_{d} $ ( it sends $ \zeta_d \mapsto\zeta_d^{d-1} $).
So I guess my question really is:
Is $ SO_8 $ the only simple Lie group whose automorphism group is not isomorphic to the automorphism group of its universal cover?
 A: Among the (connected) simple Lie groups, the following holds:

Theorem 1:  Suppose $G$ is a connected simple Lie group with universal cover $\overline{G}$ which is not of the form $Spin(4k)$.  Then $Aut(G) = Aut(\overline{G})$.

You've already indicated the argument except for $A_n$, $D_{2n+1}$, and $E_6$, but I'll give an argument which handles them all simultaneously.
The key feature in common among all these examples is:
Proposition 2:  Suppose $\overline{G}$ is a simply connected simple Lie group which is not isomorphic to $Spin(4k)$ for some $k$.  Then the center $Z(\overline{G})$ is cyclic.
Proof:  The centers are listed this website.  (If someone wants to fill in a more official reference, please do!)  $\square$
We will also need the following
Lemma 3.  Suppose $Z(\overline{G})$ is cyclic and let $H\subseteq Z(\overline{G})$ be any subgroup.  If $\overline{f}:\overline{G}\rightarrow \overline{G}$ is any automorphism, then $\overline{f}(H) = H$.
Proof:  As you already mentioned, the center of a group is always characteristic, so $\overline{f}$ restricts to a map $\overline{f}:Z(\overline{G})\rightarrow Z(\overline{G})$.  But in a cyclic group, a subgroup is uniquely characterized by its order, so $H\subseteq Z(\overline{G})$ is characteristic.  $\square$
Now we can prove Theorem 1.  I'll write $\mathfrak{g}$ for the Lie algebra of $G$.
Proof: (of Theorem 1)  Since $Aut(\overline{G})\cong Aut(\mathfrak{\overline{g}})\cong Aut(\mathfrak{g})$, it is enough to show that every automorphism of $\phi:\mathfrak{g}\rightarrow \mathfrak{g}$ comes from an automorphism $f:G\rightarrow G$.  But this follows easily from the following observations:

*

*Since $\overline{G}$ is simply connected, $\phi$ is induced by an automorphism $\overline{f}:\overline{G}\rightarrow \overline{G}$.


*$G$ is isomorphic to $\overline{G}/H$ for some $H\subseteq Z(\overline{G})$.


*The map $\overline{f}$ maps $H$ to $H$ (Proposition 2 and Lemma 3)  $\square$
I now claim that Theorem 1 is, in some sense, as good as possible.

Theorem 4  For each $k\geq 2$ (so that $Spin(4k)$ is actually simple), there is a Lie group $G$ with universal cover $\overline{G} = Spin(4k)$ for which $Aut(G)$ is smaller than $Aut(\overline{g})$.

Proof:  Set $\overline{G} = Spin(4k)$ and note that $Z(\overline{G})\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}.$  When $k=2$, you've already noted Theorem 2 is true (coming from the triality automorphism of $Spin(8)$).  So, let's assume $k\geq 3$.
Then the outer automorphism group $Out(\overline{G})$, which is isomorphic to the symmetries of the Dynkin diagram, is $\mathbb{Z}/2\mathbb{Z}$.  Let $\overline{f}:\overline{G}\rightarrow \overline{G}$ be such an outer automorphism.
As seen on page 50 of

Siebenthal, Jean de. "Sur les groupes de Lie compacts non connexes.." Commentarii mathematici Helvetici 31 (1956/57): 41-89. http://eudml.org/doc/139134,

the map $\overline{f}:Z(\overline{G})\rightarrow Z(\overline{G})$ is non-trivial.  One the other hand $\overline{f}^2$ is inner, so fixes $Z(\overline{G})$ points wise.  It follows that $Z(\overline{G})\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ has a generating set $\{a,b\}$ for which $\overline{f}$ simply swaps $a$ and $b$.
Consider the quotient of $\overline{G}$, $\pi:\overline{G}\rightarrow G:= Spin(4k)/\langle a\rangle$.
Proposition 5: There is no automorphism $f:G\rightarrow G$ which induces the same map as $\overline{f}$ on  the Lie algebra level.  That is, the automorphism $\overline{f}_\ast:\mathfrak{spin}(4k)\rightarrow \mathfrak{spin}(4k)$ is not induced by a map $f:G\rightarrow G$.
Proof:  Suppose for a contradiction that $f$ is such a map, then it must lift to $\overline{f}$ (since the two maps to the same thing on the Lie algebra level), so we obtain the commutative diagram $$\begin{matrix} \overline{G} & \xrightarrow{\overline{f}} & \overline{G}\\ \downarrow{\pi} & & \downarrow{\pi}\\ G & \xrightarrow{f} & G \end{matrix}.$$
But now start with $b\in \overline{G}$ and follow it.  We have $(\pi\circ \overline{f})(b) = \pi(\overline{f}(b)) = \pi(a) = e$, where $e$ denotes the identity element.  On the other hand, $b\notin\ker \pi$, so $\pi(b)\neq e$ and $f$ is an automorphism, so $f(\pi(b))\neq e = \pi(\overline{f}(b))$.  Thus, we have reached a contradiction.  $\square$
